If the area of an equilateral triangle is increasing at the rate of $4\sqrt{3} cm^2/sec$, then the rate of increase of its perimeter when the side is 4cm, is

6 cm/sec

The correct answer is Option (4) → 6 cm/sec

Let the side of the equilateral triangle be $x$ cm.

Area: $A = \frac{\sqrt{3}}{4}x^2$

Differentiate both sides w.r.t. time $t$:

$\frac{dA}{dt} = \frac{\sqrt{3}}{2}x \cdot \frac{dx}{dt}$

Given: $\frac{dA}{dt} = 4\sqrt{3}$ and $x = 4$

$4\sqrt{3} = \frac{\sqrt{3}}{2} \cdot 4 \cdot \frac{dx}{dt} = 2\sqrt{3} \cdot \frac{dx}{dt}$

$\Rightarrow \frac{dx}{dt} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2$

Perimeter: $P = 3x$

$\frac{dP}{dt} = 3 \cdot \frac{dx}{dt} = 3 \cdot 2 = {6}$